Origamics by Kazuo Haga

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ORIGAMICSMathematicalExplorationsthroughpaperFolding

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edited and translated by

Josefina C FonacierUniversity of Philippines, Philippines

Masami IsodaUniversity of Tsukuba, Japan

University of Tsukuba, Japan

N E W J E R S E Y L O N D O N S I N G A P O R E B E I J I N G S H A N G H A I H O N G K O N G TA I P E I C H E N N A I

World Scientific

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ORIGAMICSMathematical Explorations through paper Folding

Kazuo Haga

British Library Cataloguing-in-Publication DataA catalogue record for this book is available from the British Library.

For photocopying of material in this volume, please pay a copying fee through the CopyrightClearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission tophotocopy is not required from the publisher.

ISBN-13 978-981-283-489-8ISBN-10 981-283-489-3ISBN-13 978-981-283-490-4 (pbk)ISBN-10 981-283-490-7 (pbk)

All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means,electronic or mechanical, including photocopying, recording or any information storage and retrievalsystem now known or to be invented, without written permission from the Publisher.

Copyright 2008 by World Scientific Publishing Co. Pte. Ltd.

Published by

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ORIGAMICSMathematical Explorations Through Paper Folding

ZhangJi - Origamics.pmd 8/13/2008, 7:52 PM1

August 11, 2008 11:23 World Scientific Book - 9in x 6in Origamicsnoline

Introduction

The art of origami, or paper folding, is a great tradition in Japan. In itssimplest form, the folding is carried out on a square piece of paper toobtain attractive figures of animals, flowers or other familiar figures.The art enjoys great popularity and appeal among both young and old,and it has spread to other countries beyond Japan.

It is easy to see that origami has links with geometry. Creases andedges represent lines, intersecting creases and edges make angles, theintersections themselves represent points. Because of its manipulativeand experiential nature, origami could become an effective context forthe learning and teaching of geometry.

In this book, origami is used to reinforce the study of geometry, withthe hope that the popularity and appeal for the former will stimulatethe latter. The activities in this book differ from ordinary origami inthat no figures of objects result. Rather, they lead the reader to studythe effects of the folding and seek patterns.

The author, Dr. Kazuo Haga, is a retired professor of biology at theUniversity of Tsukuba, Japan. His interest in science has been chan-neled to the broader field of science education. He mentioned in hisbook that during his career as a biology professor, while waiting for hisexperiments to progress, he used to while away the time doing paperfolding (or more specifically, mathematics through paper folding).

The experimental approach that characterizes much of science ac-tivity (and possibly much of Professor Hagas work as a biologist) can berecognized throughout the book. The manipulative nature of origamiallows much experimenting, comparing, visualizing, discovering and

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Origamics: Mathematical Explorations Through Paper Folding vi

conjecturing. In every topic, the exuberance that the author felt when-ever he arrived at mathematical ideas is reflected in his writing style.To paraphrase the author, more wonders emerge!

Admittedly proof is a necessary part of mathematical discourse.However, proofs are not emphasized in this book. The author is awarethat many students do not appreciate formal proofs. So while someproofs are given after the paper folding, not all mathematical discov-eries are proven. The reader is encouraged to fill in all the proofs, forhis/her own satisfaction and for the sake of mathematical complete-ness.

This then is a resource book for mathematics teachers and mathe-matics teacher educators. It is hoped that going through this book willgive them alternative approaches for reinforcing and applying the the-orems of high school geometry and for provoking more enthusiasm formathematics study.

Josefina.C. FonacierFormer Director,National Institute for Science and Mathematics EducationDevelopment,University of Philippines


Until the Publication of the EnglishEdition

When I was an undergraduate student almost 30 years ago, our stu-dents mathematics research club, which aimed for understandingmathematics through different ways, held a mathematics exhibition.One of the exhibits was on Origami, paper folding: the mathemat-ics in Orizuru (crane construction), based on the work of ProfessorKoji Fushimi, a physicist, then the President of the Science Councilof Japan. We well remember that Professor Kazuo Haga visited our ex-hibition, and he explained to us the Haga theorems. He was a biologist,and we were surprised that these works on mathematics and origamihad been done by scientists (Fushimi a physicist, and Haga a biologist)and not by mathematicians. Now we are working as mathematics ed-ucators in universities, middle schools and high schools, and it was apart of important experience for us in becoming teachers.

When I came back to the University of Tsukuba 15 years ago, Pro-fessor Haga began to teach school teachers his mathematical theory ofOrigami under the name of ORIGAMICS. I recommended the pub-lisher of the Teachers Journal on Mathematics Education at MeijiTosho-Shuppan to have the serial of Professor Hagas Origamics, be-cause we knew the importance of his activity for mathematics educa-tion and teacher education. Based on the series, he published his firstbook, which would become the major resource of this English trans-lation. Since then he has published two more books. This Englishtranslation includes only one third of his works on Origamics.

There are several unique points in his Origamics. The first onecomes from the object itself. Everyone has experience in folding a pa-per, but he explored it based on his unique geometrical ideas. Another

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Origamics: Mathematical Explorations Through Paper Folding viii

point is his approaches in mathematics. He used school mathemat-ics that could be understood by anyone who has studied mathematicsat school. Through his mathematical viewpoint, we can learn how toexplore and enjoy daily situations geometrically, and develop our math-ematical views and minds in the world.

Today, through international conferences, origamics has become awell-known research field throughout the world. Some of ProfessorHagas works are lectured by himself at these conferences; at the sametime many of his works have been spread through teachers. In thecase of the Philippines, Mr. Mikio Masuda, who had been a teacherat the Junior high school (middle school) attached to the Universityof Tsukuba, was dispatched to the University of the Philippines Na-tional Institute for Science and Mathematics Education Development(UP-NISMED) as a specialist of the Japan International CooperationAgency (JICA) on the appointment of Professor Shizumi Shimizu, Uni-versity of Tsukuba. Among the people he worked with was Profes-sor Josefina C. Fonacier; she was especially impressed with ProfessorHagas work. The major part of this English edition of his book origi-nates from results of their collaborations.

Based on the experience of international cooperation with UP-NISMED through JICA, as well as other international cooperationprojects/experiences, the University of Tsukuba established the Cen-ter for Research on International Cooperation in Educational Develop-ment (CRICED) on behalf of the Ministry of Education, Japan. For de-veloping materials for international cooperation, CRICED staff mem-bers have begun fully support for publication. It is my pleasure to editthe English edition of Professor Hagas book with Professor Fonacier,on behalf of my long exchange with Professor Haga and the collabora-tion experience with UP-NISMED.

Masami IsodaCenter for Research on International Cooperation in EducationalDevelopment (CRICED)University of Tsukuba, JAPAN


Acknowledgments

We would like to acknowledge the following contributors and Institu-tions:

Dr. Soledad Ulep, Deputy Director of UP-NISMED (the Universityof the Philippines National Institute for Science and Mathematics Ed-ucation Development) for her support of our editorial works;

Dr. Yasuo Yuzawa, researcher of CRICED (Center for Research onInternational Cooperation in Educational Development, University ofTsukuba) for his major contribution for developing editorial versionfrom his mathematical expertise including pictures using Software,Cabri Geometry II+ and LATEX ;

Dr. Rene Felix, professor at the mathematics department of theUniversity of the Philippines, for his mathematical expertise and carein reading and reviewing the whole manuscript;

Mr. Mikio Masuda, retired teacher of the Junior High School at-tached to the University of Tsukuba, for his earlier assistance of trans-lation between Professors Haga and Fonacier;

Professor Shizumi Shimizu, the graduate school of human compre-hensive science, the University of Tsukuba, for his support to developinternational relationship between University of the Philippines andthe University of Tsukuba;

Ms. Foo Chuan Eng, Education Officer, Brunei Darussalam for hersupport of developing captions;

Dr. Hiroshi Yokota, researcher of CRICED for his advice of LATEX

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Origamics: Mathematical Explorations Through Paper Folding x

Programing;

JICA (Japan International Cooperation Agency) and its staff mem-bers for providing the supported to establish and develop UP-NISMEDand the opportunity to meet and share ideas with professionals fromdifferent countries; and UP-NISMED and CRICED staff members whohave given full support.

The grant for developing materials under the collaboration betweenCRICED and JICA from the Ministry of Education, Japan was used forfinalizing the part of edited version.


Preface for the English Edition

In 2001, I was given the opportunity to talk about ORIGAMI andORIGAMICS in the plenary lecture at the Third International Meet-ing of Origami Science, Mathematics and Education in Asilomar Con-ference Center, California.

ORIGAMI has become an international word at present, such asin Origami Science, which is originally derived from the Japanese wordorigami. They differ somewhat in meaning as well as pronunciation.The accent of the former falls on the third syllable (ga) while the latteron the second one (ri), that is, ori[ga]mi and o[ri]gami. Most Japanesepronounce it with a nasal sound. In Japan, origami is usually a hand-icraft hobby designed mainly for children. Thus almost all the origamibooks are in the juvenile sections of bookstores, even though some arefor enthusiasts and origami scientists. I felt the necessity to give anew name for describing the genre of scientific origami, hence I pro-posed the term ORIGAMICS at the Second Origami Science Meetingin 1994.

The term origamics is composed of the stem origami and the suffixics, which is often used to indicate science or technology, as in math-ematics. Another definite difference between origami and origamics istheir end product. The former produces paper models of animals, flow-ers, fruits, vehicles and so forth; while the latter often does not createbeautiful or skillful products, but rather some paper with a lot of wrin-kles, furrows or creases.

My first mathematical findings on origami were done in 1978. Atthat time I was a biologist majoring in arthropodan morphology, and

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Origamics: Mathematical Explorations Through Paper Folding xii

observing tiny insect embryos under a microscope, however, there wasno relation between origami mathematics and insect egg study. Asthe microscopic study needed much time with mental fatigue and eye-strain, I often had a recess and folded a piece of paper torn off froma small notebook for refreshment. Then I discovered some interest-ing phenomena in the folded paper, and corresponded about them withProfessor Koji Fushimi, who was a famous theoretical physicist andalso known as origami geometrician. He introduced my findings in themonthly magazine Sugaku (Mathematics) Seminar 18(1):40-41, 1979,titled Origami Geometry, Haga Theorem (in Japanese). The theoremwas named by him using my surname.

In the subsequent years I discovered several more phenomena onthe square and rectangular sheets of paper one after another. Thedetailed explanations were published in Japanese magazines such asSugaku Seminar (Nihon-Hyoron-Sha), Sugaku Kyoiku (Meiji Tosho-Shuppan), ORU (Soju-Sha) and Origami Tanteidan (Nihon OrigamiGakkai).

I published three books on Origamics, namely: Origamics niyoruSugaku Jugyo (Meiji Tosho-Shuppan), 1996, Origamics Part 1. Geo-metrical Origami (Nihonhyoron-sha), 1998 and Origamics Part 2. FoldPaper and Do Math (ditto), 2005. My colleagues recommended me towrite an English version of these books. Prof. Josefina C. Fonacierof University of Philippines and Mr. Mikio Masuda of University ofTsukuba (see Until the Publication of the English Edition) showedspecial interest and they eagerly drove to translate one of those booksinto English and to publish it. I responded them and started transla-tion. However, due to my retirement and changing circumstances, I didnot managed to complete it. I greatly appreciated and in debt to theirkindness and encouragements.

Fortunately, after 10 years of interruption, Associate ProfessorMasami Isoda of the University of Tsukuba proposed to make a newlyedited English version of Origamics as a part of his CRICED activities.I naturally agreed and added new chapters to the original plan. I gavemy hearty thanks to Professor Isoda for his proposal and collaborationwith Professor Fonanier.

Kazuo Haga


Contents

Introduction v

Until the Publication of the English Edition vii

Acknowledgments ix

Preface for the English Edition xi

1. A POINT OPENS THE DOOR TO ORIGAMICS 1

1.1 Simple Questions About Origami . . . . . . . . . . . . . 1

1.2 Constructing a Pythagorean Triangle . . . . . . . . . . 2

1.3 Dividing a Line Segment into Three Equal Parts Usingno Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 Extending Toward a Generalization . . . . . . . . . . . 8

2. NEW FOLDS BRING OUT NEW THEOREMS 11

2.1 Trisecting a Line Segment Using Hagas SecondTheorem Fold . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 The Position of Point F is Interesting . . . . . . . . . . . 14

2.3 Some Findings Related to Hagas Third Theorem Fold . 17

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Origamics: Mathematical Explorations Through Paper Folding xiv

3. EXTENSION OF THE HAGAS THEOREMS TOSILVER RATIO RECTANGLES 21

3.1 Mathematical Adventure by Folding a Copy Paper . . . 21

3.2 Mysteries Revealed from Horizontal Folding of CopyPaper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3 Using Standard Copy Paper with Hagas ThirdTheorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4. X-LINES WITH LOTS OF SURPRISES 33

4.1 We Begin with an Arbitrary Point . . . . . . . . . . . . . 33

4.2 Revelations Concerning the Points of Intersection . . . 35

4.3 The Center of the Circumcircle! . . . . . . . . . . . . . . 37

4.4 How Does the Vertical Position of the Point of Intersec-tion Vary? . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4.5 Wonders Still Continue . . . . . . . . . . . . . . . . . . . 41

4.6 Solving the Riddle of 1

2 . . . . . . . . . . . . . . . . . . 42

4.7 Another Wonder . . . . . . . . . . . . . . . . . . . . . . . 43

5. INTRASQUARES AND EXTRASQUARES 45

5.1 Do Not Fold Exactly into Halves . . . . . . . . . . . . . 46

5.2 What Kind of Polygons Can You Get? . . . . . . . . . . . 46

5.3 How do You Get a Triangle or a Quadrilateral? . . . . . 48

5.4 Now to Making a Map . . . . . . . . . . . . . . . . . . . 49

5.5 This is the Scientific Method . . . . . . . . . . . . . . . 53

5.6 Completing the Map . . . . . . . . . . . . . . . . . . . . 53

5.7 We Must Also Make the Map of the Outer Subdivision 55

5.8 Let Us Calculate Areas . . . . . . . . . . . . . . . . . . . 57


xv Contents

6. A PETAL PATTERN FROM HEXAGONS? 59

6.1 The Origamics Logo . . . . . . . . . . . . . . . . . . . . . 59

6.2 Folding a Piece of Paper by Concentrating the Four Ver-tices at One Point . . . . . . . . . . . . . . . . . . . . . . 60

6.3 Remarks on Polygonal Figures of Type n . . . . . . . . . 63

6.4 An Approach to the Problem Using Group Study . . . . 64

6.5 Reducing the Work of Paper Folding; One Eighth of theSquare Will Do . . . . . . . . . . . . . . . . . . . . . . . . 65

6.6 Why Does the Petal Pattern Appear? . . . . . . . . . . . 66

6.7 What Are the Areas of the Regions? . . . . . . . . . . . 70

7. HEPTAGON REGIONS EXIST? 71

7.1 Review of the Folding Procedure . . . . . . . . . . . . . 71

7.2 A Heptagon Appears! . . . . . . . . . . . . . . . . . . . . 73

7.3 Experimenting with Rectangles with Different Ratiosof Sides . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

7.4 Try a Rhombus . . . . . . . . . . . . . . . . . . . . . . . 76

8. A WONDER OF ELEVEN STARS 77

8.1 Experimenting with Paper Folding . . . . . . . . . . . . 77

8.2 Discovering . . . . . . . . . . . . . . . . . . . . . . . . . . 80

8.3 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

8.4 More Revelations Regarding the Intersections of theExtensions of the Creases . . . . . . . . . . . . . . . . . 85

8.5 Proof of the Observation on the Intersection Points ofExtended Edge-to-Line Creases . . . . . . . . . . . . . . 89

8.6 The Joy of Discovering and the Excitement of FurtherSearching . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

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Origamics: Mathematical Explorations Through Paper Folding xvi

9. WHERE TO GO AND WHOM TO MEET 93

9.1 An Origamics Activity as a Game . . . . . . . . . . . . . 93

9.2 A Scenario: A Princess and Three Knights? . . . . . . . 93

9.3 The Rule: One Guest at a Time . . . . . . . . . . . . . . 94

9.4 Cases Where no Interview is Possible . . . . . . . . . . 97

9.5 Mapping the Neighborhood . . . . . . . . . . . . . . . . 97

9.6 A Flower Pattern or an Insect Pattern . . . . . . . . . . 99

9.7 A Different Rule: Group Meetings . . . . . . . . . . . . 99

9.8 Are There Areas Where a Particular Male can have Ex-clusive Meetings with the Female? . . . . . . . . . . . . 101

9.9 More Meetings through a Hidden Door . . . . . . . . . 103

10. INSPIRATION FROM RECTANGULAR PAPER 107

10.1 A Scenario: The Stern King of Origami Land . . . . . . 107

10.2 Begin with a Simpler Problem: How to Dividethe Rectangle Horizontally and Vertically into3 Equal Parts . . . . . . . . . . . . . . . . . . . . . . . . 108

10.3 A 5-parts Division Point; the Pendulum Idea Helps . . 111

10.4 A Method for Finding a 7-parts Division Point . . . . . 115

10.5 The Investigation Continues: Try the Pendulum Ideaon the 7-parts Division Method . . . . . . . . . . . . . . 117

10.6 The Search for 11-parts and 13-parts DivisionPoints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

10.7 Another Method for Finding 11-parts and 13-parts Di-vision Points . . . . . . . . . . . . . . . . . . . . . . . . . 122

10.8 Continue the Trend of Thought: 15-parts and 17-partsDivision Points . . . . . . . . . . . . . . . . . . . . . . . 125

10.9 Some Ideas related to the Ratios for Equal-partsDivision based on Similar Triangles . . . . . . . . . . . 130


xvii Contents

10.10 Towards More Division Parts . . . . . . . . . . . . . . . 134

10.11 Generalizing to all Rectangles . . . . . . . . . . . . . . . 134

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Topic 1

A POINT OPENS THE DOOR TOORIGAMICS

Hagas First Theorem and its Extensions

1.1 Simple Questions About Origami

Whenever an origami activity is brought up in the classroom the stu-dents show great interest and enthusiasm. And even as the coloredpieces of origami paper are distributed, the students are in a hurryto start some folding process. This burst of eagerness of the studentsallows for a smooth introduction to the subject matter of this topic.

As the students make their first fold, call their attention to all thefirst folds. The objects the students plan to make may vary - flower,animal or whatever. But no matter what they are trying to make, theirfirst fold is invariably one of these types: a book fold (or side-bisectorfold) made by placing one side on the opposite side and making a creaseas in Fig.1.1(a), or a diagonal fold made by placing a vertex on theopposite vertex and making a crease as in Fig.1.1(b). In the book foldtwo opposite sides are bisected, hence the alternate name.

Why are there only two types? We explain this. In origami all ac-ceptable folds must have the property of reproducibility - the result ofa folding procedure must always be the same. The basic origami foldsinvolve point-to-point or line-to-line. Using only the four edges and thefour vertices, the possible ways of folding are placing an edge onto an-other edge or placing a vertex onto another vertex. By considering allsuch manipulations one sees that the only possible outcomes are thetwo folds mentioned above.

1


Hagas First Theorem and its Extensions 2

(a)

(b)

Fig. 1.1 The first folds with the property of reproducibility.

While musing over the above observations one might ask: whatother folds are possible if, in addition to the four vertices, another pointon the square piece of paper were specified? This question plants theseed for the discussions in this book and opens the door to origamics -classroom mathematics through origami.

1.2 Constructing a Pythagorean Triangle

When we are told to select a particular point on the square paper otherthan the vertices without using any tool (that is, no ruler or pencil), thesimplest to be selected is the midpoint of a side. To mark a midpointstart bending the paper as for a book fold, but do not make a full crease.Just make a short crease on the edge of the square or make a shortmark with ones fingernails. It is not necessary to make a crease thewhole length of the paper; too many crease marks are likely to be anobstacle to further study. We shall call a small mark like this a scratchmark or simply a mark.

Now we make a fold on the paper with this midpoint as reference orstarting point. Several methods of folding can be devised. One foldingmethod is to place a vertex on the mark, another folding method is tomake a crease through the mark. The method should be such that a


3 A POINT OPENS THE DOOR TO ORIGAMICS

Fig. 1.2 Make a small mark on the midpoint of the upper edge.

Fig. 1.3 Place the lower right vertex onto the midpoint mark.

unique fold is obtained, no matter how often or by whom it is made.

In this topic we shall discuss one folding procedure and some prop-erties related to it. Other ways of folding shall be discussed in othertopics.

To facilitate discussion let us set the standard position of the squarepiece of paper to be that where the sides are horizontal (that is, left toright) or vertical (that is, upwards or downwards). Therefore we shalldesignate the edges as left, right, upper, or lower; and the vertices asupper left, lower left, lower right, or upper right.



Select the midpoint of the upper edge as starting point (Fig.1.2).Place the lower right vertex on the starting point and make a firmcrease (Fig.1.2). Either the right or left lower vertex may be used, itdoes not make any difference for analysis purposes. But to follow thediagrams we shall use the lower right vertex.

By this folding process a non-symmetrical flap is made. A numberof interesting things can be found about it. To facilitate discussion, inFig.1.4 points were named.

A

B C

DE

F

G

IJ

H

Fig. 1.4 There are three similar right angled triangles.

Let the length of one side of the square be 1.

First, in right 4DEF we can find the lengths of the sides. Let DF =a. Then FC = 1 a. By the folding process FE = FC, so FE = 1 a.Since E is a midpoint, DE =

1

2. Applying the Pythagorean relation,

(1 a)2 = a2 +(

1

2

)2. From this we obtain a =

3

8. Therefore DF =

3

8

and FE= 1 a = 58

. In other words by the above folding procedure theright side of the square is divided in the ratio 3 : 5. And further, theratio of the three sides of 4EDF is

FD : DE : EF =3

8:

1

2:

5

8= 3 : 4 : 5.

4EDF turns out to be a Pythagorean Triangle!Such triangles were used by the Babylonians, the ancients Egyp-

tians such as for land surveying along the lower Nile River and the an-



cients Chinese. History tells us that several thousand years ago therewas repeated yearly flooding of the river, so land boundaries were con-tinually erased. For resurveying these boundaries they made use of thePythagorean triangle. The 3 : 4 : 5 triangle is often mentioned as theorigin of geometry.

Constructing the Pythagorean triangle by Euclidean methods - thatis, with the use of straight edge and compass - requires a lot of time.By contrast, as you have seen, this can be done in origamics with justone fold on the square piece of paper.

1.3 Dividing a Line Segment into Three Equal Parts Us-ing no Tools

Still other triangles emerge from the folding procedure. The lengths oftheir sides reveal some interesting things.

We determine the lengths of the sides of4EAH in Fig.1.4. As before,let the length of the side of the square be 1. Since vertex C of the squarewas folded onto point E and C is a right angle, then also HEF is a rightangle. Therefore the angles adjacent to HEF are complementary and4EAH and 4FDE are similar. Therefore 4AEH is also an Egyptiantriangle.

Now we look for AH. By the proportionality of the sides

DF

DE=

AE

AH, then

3

81

2

=

1

2AH

.

Therefore AH =2

3.

This value of AH is another useful surprise. It indicates that by

locating the point H one can find1

3of the side - BH is

1

3of the side.

That is, H is a trisection point.

Dividing a strip of paper into three equal parts is often done bylightly bending the strip into three parts and shifting these parts ina trial-and-error fashion until they appear equal. Because trial-and-error is involved this method is imprecise and therefore is not mathe-



matically acceptable. Other trisecting methods by origami have beenreported, but the method above described is one of the simplest andneatest. In fact, it is possible to carry out the procedure of marking thetrisection point with almost no creases.

We continue to look for the other sides of 4AEH. We look for sideHE.

DF

EF=

AE

HE, then

3

85

8

=

1

2HE

.

Therefore HE =5

6.

This value of HE is also useful in that it enables us to find1

6of the

side. By returning the flap to the original position EH falls on side CB,

so that H separates1

6of the side. That is, H is a hexasection point of

the side.

There is still another triangle to study in Fig.1.4, right angled4GIH. Since GHI and EHA are vertical angles and are thereforeequal, then 4GIH and 4EAH are similar. So 4GIH is still anotherEgyptian triangle with

GI : IH : HG = 3 : 4 : 5.

Also, since EI = CB = 1, then HI = EI EH = 1 56

=1

6. As for the

other sides of 4GIH, it is easy to obtain GI = 18

and GH =5

24.

Finally, to complete our study of the segments in Fig.1.4, we lookfor the length of FG. Imagine a line (or fold) through G parallel to thelower edge BC and intersecting side CD at point J. This line forms aright 4FJG with hypotenuse FG. Since by folding GB = GI, then GI= JC =

1

8and JF = CF CJ = 5

8 1

8=

1

2. Therefore by applying the

Pythagorean Theorem to 4FJG, FG =

5

2.

The main ideas just discussed are summarized in the following the-orem.



Hagas First Theorem By the simple folding procedure of plac-ing the lower right vertex of a square onto the midpoint of the upperside, each edge of the square is divided in a fixed ratio, as follows (seeFig.1.5 ).

(a) The right edge is divided by the point F in the ratio 3 : 5.(b) The left edge is divided by the point H in the ratio 2 : 1.(c) The left edge is divided by the point G in the ratio 7 : 1.(d) The lower edge is divided by the point H in the ratio 1 : 5.

1

2

5

6

5

8

1

24

56

1

8

58

3

2

1

3

2

8

5

2

1

Fig. 1.5 Various lengths appears by folding only once.

And the fold used in the theorem is called Hagas First TheoremFold.

In (a) and (c) the ratios may be obtained by dividing a side in half,then again in half, then still again in half (that is, dividing the side into8 equal parts). But the ratios in (b) and (d) cannot be so obtained. Forthis reason this one-time folding method is a useful, simple and precisedividing procedure.

Comment. The discoveries described in this topic were first reportedas Hagas Theorem by Dr. Koji Fushimi in the journal MathematicsSeminar volume 18 number 1 (January 1979, in Japanese). Other fold-ing methods have since been explored by Haga, hence the change inname in 1984 to Hagas First Theorem.



Dr. Fushimi is a past chairman of the Science Council of Japan.He is author of Geometrics of Origami (in Japanese) published by theNippon Hyoronsha.

1.4 Extending Toward a Generalization

So far the folding procedures have been based on the midpoint of anedge as starting point. We might ask ourselves: what results would weobtain if the starting point were some other point on the edge?

In Fig.1.6 an arbitrary point was chosen and is indicated by anarrow. In Fig.1.7 the vertices of the squares are named. Denote thechosen point by E and the distance DE by x. Denote the differentsegments by y1 to y6 as in Fig.1.7. Then the lengths of the segmentsbecome functions of x as follows.

Fig. 1.6 Folding onto positions other than the midpoint.

[y1] By the Pythagorean relation on 4DEF, x2 + y21 = (1 y1)2. Soy1 =

1 x22

=(1 + x)(1 x)

2.

[y2] Since 4AHE is similar to 4DEF, then y11 x =

x

y2. So

y2 =2x

1 + x.

[y3] Also from similar triangles 4AHE and 4DEF, we obtainy2

y3=

x

1 y1 . So y3 =1 + x2

1 + x.



3

65

4

3

2

1

1

1

1

y

yy

y

yy

y

y

1 x

4y

1

11 y

xA

B C

DE

F

G

HI

J

K

Fig. 1.7 y1 to y6 indicates the other various lengths.

[y4] Since FG and EC are perpendicular, 4CKF and 4CDEare similar. Therefore DEC and KFC are congru-ent; so also 4CDE and 4GJF. So FJ = x. Thereforey4 = JC = 1 (y1 + x) = (1 x)

2

2.

[y5] Since y2 + y5 + y4 = 1, y5 = 1(

2x

1 + x+

(1 x)22

).

[y6] By the Pythagorean relation on 4GJF, y6 =

FJ2 + JG2 =x2 + 1.

It is difficult to feel excited over the above relations if described onlyin terms of formal general expressions. To help us better appreciatethese relations let us find their values for particular values of x. Using

the square pieces of paper, locate the points corresponding to x =1

4and

x =3

4. We fold as before, placing the lower vertex on each mark as in

Figs.1.8(a) and (b).

The values of the ys for these two values of x, as well as those for

x =1

2, are given in the table below.

From the table we see that various fractional parts are produced,the simpler ones being halves, thirds, fourths, fifths, sixths, seventhsand eighths. We realize that by selecting suitable values of x we canobtain segments of any fractional length or their integral multiples.Therefore with no tools, by simply marking a specific dividing pointon an edge and making just one fold, any fractional part of the square



(a)

Case: x =1

4

(b)

Case: x =3

4

Fig.1.8

piece of paper may be obtained. And to reduce the clutter of too manyfolds, just fold in parts (i.e., small scratch marks instead of wholecreases) to obtain the important points.

x

y1

3

8

5

8

15

32

17

32

7

32

25

32

2

3 1

3

2

5

3

5

6

7

1

7

5

6

1

6

17

20

3

20

25

28

3

28

1

8

7

8

9

32

23

32

1

32

31

32

1

2

1

4

3

4

y2

y3

y4

1 y1

1 y2

1 y3

1 y4

Thus, in spite of the austere simplicity of this one-fold procedure,many exciting revelations emerge. Clearly Hagas First Theorem Foldis highly worthwhile.


Topic 2

NEW FOLDS BRING OUT NEWTHEOREMS

Mathematical Principles Related to Hagas Second and ThirdTheorems

2.1 Trisecting a Line Segment Using Hagas SecondTheorem Fold

In the previous topic, just by placing a lower vertex of a square piece ofpaper on the midpoint of the upper edge and making a crease many in-teresting ideas about the resulting segments and angles came to light.One such idea was discussed in topic 1 as Hagas First Theorem, andthe related fold was named Hagas First Theorem Fold (Fig.2.1(a)).

In the present topic we shall discuss Hagas Second Theorem Fold.As before mark the midpoint of the upper edge of the paper. Thenmake the fold linking this midpoint and a vertex of the bottom edge.This is an unusual way of folding and you may find it a bit difficult;so fold very carefully, especially when folding through the midpoint(Fig.2.1(b)). Make a firm crease.

In Fig.2.2 points were named: E is the midpoint of side AD of thesquare ABCD. EC is the resulting crease and right 4EFC is the re-sulting triangular flap. If the length of a side of the square is 1, then

the length of the crease EC is

5

2, obtained by using the Pythagorean

theorem on right 4EDC.We shall study the folded part of the top edge, EF in Fig.2.2. This is

the new position of ED after folding. Suppose segment EF is extendedto the left edge at point G. Where does this extended line reach on the

11


Mathematical Principles Related to Hagas Second and Third Theorems 12

Fig. 2.1 How to fold the upper edge through the midpoint?

Fig. 2.2 Where is the position of G, when the extended line EF meets line AB?

left side?

When I folded this way the first time, I intuitively sensed that thepoint G would be a trisection point of the left side (because I was study-ing this right after I found the trisection point by Hagas First Theoremfold).

Comment on teaching. On a few occasions I asked several of my stu-dents the following question:

When a piece of origami paper is folded as shown inFig.2.2, and the line EF is extended to meet the left side


13 NEW FOLDS BRING OUT NEW THEOREMS

at point G, it seems that the length of BG is one third ofthe left side. Is this true?

It took some my college students (biology majors) a long time to answerthe question; others gave up. It took some middle school students lesstime to solve this. One entomologist from a foreign country who wasvisiting Japan sent me his answer one month after his return to hishome country.

To help us answer the question fold the paper once again as inFig.2.3. That is, place the bottom edge BC on the edge EC of the trian-gular flap to make a second triangular flap. Since these two edges aresides of the square paper their lengths are the same, and since bothadjoining vertices are right angles the two short legs of the trianglesare collinear. Therefore the end of the fold is point G.

Fig. 2.3 Two pair of similar right angled triangles are used to prove G is a trisectionpoint.

Denote the areas of the triangular flaps as R and S respectively, andthe area of4AEG as T. Then the area of the whole square is 2R+2S+T.Assuming that the length of one side of the square is 1 and letting BG= x, then

R =1

4, S =

x

2, T =

1 x4

.

Therefore the expression for the area of the square becomes

2

(1

4

)+ 2

(x2

)+

1 x4

= 1.



From this equation we obtain x=1

3.

So we made a good guess that the point G is a trisection point.

There are other ways to prove that G is a trisection point; thinkof other proofs. A proof is possible without use of the Pythagoreantheorem. Constructing a second proof could be an exercise for secondgrade middle (8th grade) students.

Hagas Second Theorem Mark the midpoint of the upper edgeof a square piece of paper, and make a crease through this midpointand the lower right vertex. A right triangular flap is formed. If theline of the shorter leg of the flap is extended to intersect the left edgeof the square, the intersection point divides the left edge into two

parts, the shorter part =1

3of the whole edge (Fig.2.2).

Comment. This theorem was introduced by Mr. Yasuhari Hushimi asHagas Second Theorem in the extra issue Origami no Kagaku (Sci-ence of Origami) attached to the Journal of Science, October 1980, pub-lished by the Nihon Keizai Shinbun-sya. And the fold produced wasnamed the Second Theorem Fold.

2.2 The Position of Point F is Interesting

The position of the point F may be arrived at by using ratios of corre-sponding sides of similar triangles.

See Fig.2.4. Let the intersection of DF and EC be H, and let I be thefoot of the perpendicular from F to the upper edge. The right 4CDEand 4DHE have a common acute angle, and right 4DHE and 4DIFhave a common acute angle, therefore the three triangles are similar 1.

From 4CDE 4DHE, DH= 15

. So DF= 2DH =25

.

From 4CDE 4DIF, CD : CE = DI : DF.Then 1 :

5

2= DI :

25

. So DI =4

5.

1Here, Similar is represented by the symbol .



Fig. 2.4 Various lengths could be obtained by the Second Theorem fold. Various lengthscould be obtained by the Second Theorem fold.

Therefore AI=1

5; that is, point F is

1

5from the left edge.

From 4CDE 4DIF, DI : FI = 2 : 1. So FI= 25

; that is, point F is =2

5from the upper edge.

Comment. This property was discovered by Mr. Kunihiko Kasahara.

Thus, by just marking the midpoint of the upper side of the square

paper and making a Second Theorem fold, one can easily obtain1

5,2

5,3

5

or4

5of the side of the square. You cannot obtain these results as easily

by using compass and straightedge.

Comment on teaching. As a classroom activity, how a teacher devel-ops the topic depends on his/her style. However, before bringing up amathematical proof it would be more concrete and interesting for thestudents if a folding procedure to support the above findings is broughtup first. Following is a suggestion. The point F is first located by HagasSecond Theorem Fold. The teacher then proposes that the position ofF is as stated above. Then he/she asks for paper folding procedures tosupport the proposition. Finally, after paper folding, he/she develops aproof.



A sample folding procedure to verify that F is1

5from the left side

is as follows (Figs.2.5(a) and (b)). On a piece of square paper mark thepoint F (Hagas second theorem fold).

(1) Place side CD onto F to make a vertical fold. This makes arectangular flap. Unfold.

(2) Place side CD onto the vertical fold of step (1). This makes arectangular flap and a second vertical fold. Unfold.

(3) Move side AB onto the vertical fold of step (2). This makes arectangular flap and a third vertical fold. Unfold.

(4) Divide the rectangular flap of step (3) into two equal parts tomake a fourth vertical fold. This last fold should pass throughF.

F F

F

F

A

B C

D A

B C

D

A

B C

D A

B C

D

(a) (b)

(c) (d)

Fig. 2.5 5 equal division of the paper can be produced by using point F.

A sample folding procedure to show that F is2

5from the top edge is

shown in Fig.2.5(c) and (d).

The folding procedures may be repeated with 2 or 3 other pieces ofpaper to be more certain of the results. Finally before the end of the



lesson a mathematical proof should be developed.

2.3 Some Findings Related to Hagas Third Theorem Fold

Two folds have been brought up of folding a square piece of paper usingthe midpoint of the upper edge as reference point. There is still anotherway of folding.

Comment. I thought of this folding more than 10 years after the pub-lication of Hagas Second Theorem. My profession is biology, and for awhile I concentrated on a new phenomenon in my field, and so I did nothave time to play with those paper squares. I used to do paper foldingwhile riding on a bus or train. I thought of this new way of foldingwhile I was on a bus going from Tokyo station to Tsukuba Center. I didnot notice that I was talking to myself at that time, until some of theother passengers started staring at me. Suddenly I felt embarrassed.Nevertheless, I was excited over my discovery and continued to look forrelations as the bus proceeded.

Fig. 2.6 There is yet another way of folding the paper onto the midpoint of the upperedge.

The new folding procedure is shown in Fig.2.6. The starting point isalso the midpoint of the upper edge. Mark this midpoint. Then lightlybend the paper so that the right vertex falls on the left edge; do not



make a crease. Shift this lower vertex along the left edge until theright edge is on the marked midpoint. Hold this position and make afirm crease. Remember: the lower vertex should be on the left edge andthe right edge should pass through the marked midpoint.

Fig. 2.7 The ideas of the clarification of the Third Theorem fold.

It appears that the intersection on the left edge cuts off1

3of the

edge. We proceed to prove that this is indeed true.

With the help of Fig.2.7 we shall prove that HB =1

3.

Assume that the length of the side of the square is 1. Let HB = xand BG = y. By the paper folding procedure

CG = GH = 1 y.Applying the Pythagorean theorem to 4HBG,

x2 + y2 = (1 y)2,and therefore

y =1 x2

2. (1)

Since EHG is right, then AHE and BHG are complementary.Therefore 4EAH and 4HBG are similar, and AE : AH = HB : BG,or

1

2: 1 x = x : y.

This results in

y = 2x(1 x).



Substituting in (1) we obtain1 x2

4= x x2,

which leads to

3x2 4x + 1 = 0 or (3x 1)(x 1) = 0.

The roots are x = 1 and x =1

3. The value x = 1 is discarded, so

x =1

3.

Fig. 2.8 By the Third Theorem fold, not only1

3but also the segments of

1

6and

1

9are

possible.

Comment. It was fortuitous that in 1994 the Second World Confer-ence on the Science of Origami was held. At this conference I gavea talk on the opening day wherein I announced this new theorem.And I named it Hagas Third Theorem, following the First and Secondtheorems.

Hagas Third Theorem Mark the midpoint of the upper edgeof a square piece of paper. Bend the paper to place the lower rightvertex on the left edge, then shift it upward or downward until theright edge of the paper passes through the marked midpoint. Makea firm crease. The crease formed divides the left edge into two parts,

the shorter part1

3of the whole edge.



In the course of the proof I solved for the lengths of the parts of allthe sides. Figure 2.8 shows these lengths. Notice that we get lengths of,not only thirds, but also sixths and ninths. You can see how easily thesefractional parts were obtained by paper folding. By contrast, thesefractional parts are not easy to arrive at if you use straight edge andcompass only.


Topic 3

EXTENSION OF THE HAGASTHEOREMS TO

SILVER RATIO RECTANGLESVertical and Horizontal Layouts

3.1 Mathematical Adventure by Folding a Copy Paper

Square paper origami is easily available, easy to become familiar with,and easy to use; and because of this one somehow feels that squareorigami is play. The author therefore decided to use another kind ofpaper, also used frequently in everyday life. Copy paper, note paper,writing paper, report paper, memo notepads, and the like may havesides in the ratio 1 :

2. This ratio is important because when such

rectangles are folded in half, the resulting all display the same ratio1 :

2.2

Size A4 paper is popular in offices. Take a sheet of A4 paper. Toverify that the sheet has the required ratio of sides, fold a square on oneend of the rectangle with side the short end of the rectangle. Then foldthe square to obtain the diagonal (note that the diagonal has length

2).

The long side of the rectangle should be the same length as the diagonalof the square. There are two ways to position the rectangular sheet ofpaper. One way is when the long side is in the vertical position as inFig.3.1; another is such that the long side is horizontal as in Fig.3.4. We2Most countries have adopted the international standard for paper sizes, paper with

ratio of sides 1 :

2. An important property of such standard rectangles is that if youdivide a rectangle crosswise into two equal parts the resulting rectangles will also havethe same ratio of sides. Thus for the sequence of A sizes, A0, A1, A2, A3, A4, . . . each sizeis half the preceding size in area but all have ratios of sides 1 :

2. Rectangles with this

ratio of sides are also called silver rectangles.

21


Vertical and Horizontal Layouts 22

Fig. 3.1(a) Rectangular paper in a vertical layout,(b) Extension of Hagas first theorem, and(c) Extension of Hagas second theorem

shall call these paper positions the vertical and the horizontal layoutrespectively. First place the rectangle in vertical position as in Fig.3.1.Let us now construct Hagas First Theorem Fold. Mark the midpointof the top edge and fold the paper so that one vertex of the lower sidecoincides with this midpoint (the lower right vertex in Fig.3.1(b)).

We can now calculate different lengths. Be aware that the sides ofthe rectangle have different lengths so that a calculated length may bea fractional part of the short side or a fractional part of the long side.In Fig.3.2 (and the other figures in this topic), fractions referring to apart of the long side are in italics; fractions referring to the short side

are in ordinary print. For example, segment HE is9

14of the short side;

EF is9

16of the long side or

9

16of

2.

The calculations are left to the reader; the results are shown inFig.3.2. From this figure note that


23 EXTENSION OF THE HAGAS THEOREMS TO SILVER RATIO RECTANGLES

In the following figure, fractions in italics,b

a, represent the ratio

when the long side is one unit.

Fig. 3.2 Lengths obtained by Hagas first theorem of rectangle in a vertical layout.

FD is7

16of CD, or CD is divided in the ratio 7 : 9 by G;

AF is11

16of AB, or AB is divided in the ratio 11 : 5 by F;

AH is2

7of AB, or AB is divided in the ratio 2 : 5 by H;

and furthermore, for the bottom side, which was folded backto the position IE,

BC is divided in the ratio 5 : 9 by H.

Thus, points F, G and H can be used to divide the rectangle into16ths, 9ths and 14ths respectively.

Among these four ratios, of particular importance are the last two.The first two ratios imply division into 16 parts, such division easilyobtained by folding into two equal parts, then again into two equalparts, repeatedly folding again into two equal parts to finally obtain16 equal parts. But the relationships displayed by the last two implydivision into 7 or 14 equal parts; and such divisions not obtainable byfolding in any simple manner.



The calculations are left to the reader; the results are shown inFig.3.2. From this figure note that

FD is7

16of CD, or CD is divided in the ratio 7 : 9 by G;

AG is11

16of AB, or AB is divided in the ratio 11 : 5 by F;

AH is2

7of AB, or AB is divided in the ratio 2 : 5 by H;

and furthermore, for the bottom side, which was folded backto the position IE,

BC is divided in the ratio 5 : 9 by H.

Thus, points F, G and H can be used to divide the rectangle into16ths, 9ths and 14ths respectively.

Fig. 3.3 Lengths according to point F of a rectangle in a vertical layout using Hagassecond theorem.

Among these four ratios, of particular importance are the last two.The first two ratios imply division into 16 parts, such division is easilyobtained by folding into two equal parts, then again into two equalparts, repeatedly folding again into two equal parts to finally obtain16 equal parts. But the relationships displayed by the last two implydivision into 7 or 14 equal parts; and such divisions is not obtainableby folding in any simple manner.



Next, let us construct Hagas Second Theorem Fold. Make the foldconnecting the midpoint of the top edge and one lower vertex (the rightlower vertex in Fig.3.1 (c)).

We can now calculate the different lengths; see Fig.3.3. Again, bereminded that fractions referring to parts of the long side are in italics;fractions referring to the parts of the short side are in ordinary print.

Again the calculations are left to the reader. The lengths of seg-ments calculated are as follows:

AG is1

9of the short side that is, short side AD is divided in

the ratio 1 : 8 by G;

AI is2

9of the long side that is, long side AB is divided in the

ratio 2 : 7 by I;

AJ is1

4of the long side that is, long side AB is divided in

the ratio 1 : 3 by J;

AK is2

7of the long side that is, long side AB is divided in

the ratio 2 : 5 by K.

Thus the segments other than AJ also imply odd-numbered divi-sions; that is, the points G, I and K can be used to divide the rectangleinto 9 or 7 equal parts. Such divisions are extremely hard to obtain byfolding.

3.2 Mysteries Revealed from Horizontal Folding of CopyPaper

Naturally, in a discussion of one-fold paper-folding with copy paper, onemust consider both the vertical and horizontal layouts (see Fig.3.4).

First we will perform Hagas First theorem fold. Placing the 1 :

2

rectangle in horizontal layout (Fig.3.4), we mark the midpoint on thetop edge. We then fold the paper so that one vertex of the lower side(right vertex in the figure) coincides with the midpoint.

Unlike the folding for vertical layout, the triangular flap does not



Fig. 3.4(a) Rectangular paper in a horizontal layout,(b) Extension of Hagas first theorem, and(c) Extension of Hagas second theorem.

In the following figure, fractions in italics,b

a, represent the ratio

when the long side is one unit.

Fig. 3.5 Extension of Hagas first theorem fold of rectangle in a horizontal layout.

reach the left side (Fig.3.4(b)) and therefore initially the author hadlittle interest in this horizontal layout. But on more careful study ofhis folding a revelation was awaiting him. When the paper is foldedalong crease FG, the segment BF remains on the bottom side and thesegment DG remains on the right side. And this is the surprise: two

1

4 segments appear, BF is

1

4of the long side, and GD is

1

4of the short

side of the rectangle.



The reader is invited to verify this with calculations; the auxiliarylines in Fig.3.5 may be useful. Here HF is parallel to AB.

Fig. 3.6 A division point is formed in a rectangle in horizontal layout.

One other observation concerns the areas. Since HF is parallel to

AB and BF is1

4of the long side of the rectangle,

the area of the rectangle ABFH is1

4of the area of the original rectangle;

the area of4EFH is half of the area of rectangle ABFH or 18

of the areaof the original rectangle;

and the area of 4DEG is 18 1

2or

1

16of the original rectangle.

After subtracting these areas from the rectangle, half of the remain-

der is the area of 4EFG, or 932

of the rectangle, roughly between1

4and

1

3of the rectangle. This poses an interesting tidbit for the reader.

Comment. Recently while traveling on the Shinkansen Superexpress(Bullet train), I had a mysterious feeling as I performed Hagas SecondTheorem Fold with 1 :

2 paper. Because it is my habit to do paper fold-

ing when using public transportation, people sometimes turned theirheads and cast pitying eyes on me; but because of my strong concen-tration at these times, their looks did not bother me.

On one particular occasion, still on Hagas Second Theorem Fold,



I was unmindful of the skeptical glances of the other passengers, andmy attention was fixed on the new position of the vertex of the tri-angular flap. I marked this new position with a black dot (Fig.3.6).Then in an experimental frame of mind I made folds through this dotparallel to the edges of the rectangle, two more parallel to the bottomedge and equidistantly placed, and two more parallel to the side edgesand equidistantly placed. It appeared that the horizontal folds dividethe rectangle into three smaller rectangles of equal area (Fig.3.6(b)),and the vertical folds divide the rectangle into three smaller rectan-gles (Fig.3.6(c)). And a discovery dawned: the black dot seems to be astarting point for dividing the sides of the rectangle into three equalparts!

This was a big surprise for me. I have not yet been able to usethis discovery in my classes, but I am sure that it will also surprisemy students. Surprise and mystery are the motives for science,and tasting the emotion that accompanies such elucidations lays theground for succeeding discoveries.

Fig. 3.7 Verification of the 3 equal parts using Hagas second theorem fold for a rectan-gle in horizontal layout.

Now let us prove the above result. In Fig.3.7 the intersection of DHand the crease CF is called G, and the foot of the perpendicular from Fto side AD is called H.

Because the ratio of the sides of a 1 :

2 rectangle remains the samewhen the rectangle is divided crosswise into two equal parts, then therectangle with sides CD and DE and diagonal CE is also a 1 :

2 rect-



angle. And therefore, for the sides of 4EDC,

ED : DC : CE = 1 :

2 :

3.

Because the following right triangles mutually share one angleother than the right angle,

4EDC 4EGD 4FHD.

Assuming that the short side of rectangle ABCD is 1, then the long

side is

2 and for 4EDC, CD = 1, ED =

2

2, CE =

6

2.

So, from 4EGD 4EDC, DG = 13

. From 4FHD 4EDC, we

obtain FH =2

3; or FH is

2

3of the short side DH =

2

2

3, or DH is

2

3of

the long side.

The last two statements support the observation that F can be usedto divide the rectangle both vertically and horizontally into 3 equalparts.

Fig. 3.8 The Third Theorem fold with A4 paper in a horizontal layout.



3.3 Using Standard Copy Paper with Hagas ThirdTheorem

We shall discuss both the vertical and horizontal positions of the paper.

For the horizontal layout, first locate the midpoint of the upper sideand shift the lower right vertex along the left edge. You will realizethat the right edge cannot meet the midpoint of the upper side.

However, if, instead of using the midpoint you take the point mark-ing the left fourth of the upper edge and make a Third Theorem fold(that is, slide the lower right vertex along the left side), you will seethat the right edge can pass through that point (Fig.3.8). It appearsthat the point of intersection of the fold with the upper edge cuts off asegment half the length of the shorter side; does it?

For the vertical layout - that is, the long side of the rectangle is avertical, Third Theorem fold is possible through the midpoint of theupper edge (Fig.3.9). Look at the position of the shifted vertex on the

left edge; that is, point H in Fig.3.10. It looks like it cuts off1

7of the

long side. Let us find out. Let the length of the short side of the paperbe 1. Then the long side has length

2. Let HB = x and BG = y. By the

Pythagorean Theorem on 4HBG,

x2 + y2 = (1 y)2.

And therefore

y =1 x2

2. (1)

Since 4EAH and 4HBG are similar,

HB : BG = AE : AH, or x : y =1

2:(

2 x)

. (2)

Equations (1) and (2) together lead to the quadratic equation,

3x2 4

2x + 1 = 0. Therefore, x =2

2

5

3.

Discarding the positive sign and choosing the negative sign we ob-tain x = 0.19745 . . . .



Fig. 3.9 Hagas Third Theorem fold with A4 paper in a vertical layout.

Fig. 3.10 Mathematical principles of Hagas Third Theorem fold with A4 paper in avertical layout.

Comment. When I carried out this investigation I intuitively felt thatx must be

1

7of the longer side. The value of two sevenths obtained

in Hagas First theorem fold on a rectangular sheet of paper made me



think this way. I revised the calculation by letting the length of the

long side of the rectangle be 1 (and therefore the short side is12

).

Still the value of x did not come out as I had hoped. So you see, somepropositions may agree with intuition and others may not. But throughmathematics we can judge intuitive statements very clearly as right orwrong.

There are many other discoveries that can be unearthed from car-rying out Hagas First, Second and Third Theorem folds on A4 paper.I did not discuss then here, but I urge you to carry out your own in-vestigations. And finally, try to generalize your findings to any sizerectangular paper with sides in the ratio a : b.


Topic 4

X-LINES WITH LOTS OFSURPRISES3

Mathematical Ideas Related to Certain Creases Made with Respect toan Arbitrary Point

4.1 We Begin with an Arbitrary Point

In the preceding topics we started with the midpoint of an edge of asquare piece of paper. This time we mark any point on the upper edge.

Many students would probably be bewildered by this instruction.They are more accustomed to following more definite instructions, notselecting any point.

The folding method is as follows:

(a) Start with a square piece of paper. Mark any point on the upperedge of the paper. While the midpoint or endpoint is acceptable,it would be more in keeping with the term arbitrary to picksome other point.

(b) Place one lower vertex on the selected point and make a firmcrease similar to the First Theorem fold (Fig.4.1(b)). Press thewhole crease repeatedly with the finger nail to make it clearand distinct.

(c) Unfold (Fig.4.1(c)). Make sure that the fold is distinct.(d) Now place the other lower vertex on the selected point and3These surprises come from the mathematical reasoning arise through paper folding.

Through those surprises, we can find the invariant and it initiates us to inquire muchmore paper folding as mathematical science activity.

33


Mathematical Ideas Related to Certain Creases 34

make another firm, distinct crease (Fig.4.1(d)).(e) Unfold again (Fig.4.1(e)). Do you see the two creases?

!" "#!$% &

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.

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8!;@A8 !" ", B

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35 X-LINES WITH LOTS OF SURPRISES

4.2 Revelations Concerning the Points of Intersection

We shall call the pair of creases obtained above as X-creases. Withother pieces of paper make other X-creases with other starting points.Use different pieces of paper; if just one piece is used, the many creaseswould make things messy and judgments would be difficult. Besides,you will use the different pieces of paper later. For each piece of papermark the starting point with a pen or pencil.

Different X-creases are obtained by different starting points, andtherefore the position of the point of intersection may vary.

Take one piece of paper. Make a vertical book fold to obtain thevertical midline of the square. Do likewise with your other X-creases.Observe that, regardless of the starting point, the intersection falls onthe midline (Fig.4.2).

Fig. 4.2 By making a vertical book fold, it is confirmed that the intersection falls on themidline.

We state our observation:

(1) The points of intersection of the X-creases fall on the ver-tical midline.

A mathematical argument for observation (1) shall be discussedlater.

Another matter to be studied is the vertical position of the intersec-tions. From your folding you can see that the intersections lie a littlebelow the center of the square.



(a) (b)

(c) (d)

Fig. 4.3 The intersection of the X-creases changes when the position of the arbitrarypoint is changed.

Fig. 4.4 The position of the intersection varies up and down when A, B, C, and D ofFig.4.3 are brought together.

Pile up the pieces of paper which you used to make X-creases, andhold the pile up to the light. You will see that the points of intersectionseem to vary up and down along the midline, although within a smallrange (Fig.4.4).


(2) The points of intersection of the X-creases lie along themidline and lie below the center of the square within a certain range.



Fig. 4.5 The arbitrary point, lower right vertex and lower left vertex are joined to theintersection point.

Again observation (2) needs to be mathematically supported, but weshall take care of this later. We describe other interesting findings.

Select a starting point and the corresponding X-creases. Draw aline from the intersection point to the starting point. Also draw linesfrom the intersection point to the lower vertices of the square (Fig.4.5)(instead of drawing it is possible to make creases for these spokes-likelines if you prefer). Then fold along an X-crease and hold the paper upto the light. It comes out that two of the spokes coincide. Repeat withthe other X-crease. It appears that the third spoke also has the samelength.


(3) The distances from the point of intersection to the startingpoint and to each of the lower vertices are equal.

4.3 The Center of the Circumcircle!

We shall now demonstrate observations (1) and (3). Refer to Fig.4.6.

First we prove observation (1).

Draw straight lines connecting the starting point to each of thelower vertices; that is, EB and EC. This forms triangle EBC. Fold againalong one of the X-creases and hold the paper up to the light. You willsee that the line you just drew is folded into two equal parts, one partlying on the other.



That is, the X-crease is the perpendicular bisector of one of the linesyou just drew. Likewise, the other X-crease is the perpendicular bisec-tor of the other line.

Fig. 4.6 The intersection point, J, of X-creases is the circumcenter of 4EBC (the centerof circumcircle).

Recall that in a triangle the perpendicular bisectors of the sidesintersect at a common point. Since J is the intersection of the per-pendicular bisectors of sides EB and EC, then J should also lie on theperpendicular bisector of side BC - that is, the vertical midline. Thusstatement (1) has been proved.

Recall further that the intersection of the perpendicular bisectorsof the sides of a triangle is equidistant from the vertices; that is, itis the center of the circle containing the vertices of the triangle. Inother words, J is the center of the circumcircle of 4EBC. This provesstatement (3).

Certainly these paper folding experiences could arouse in the stu-dents greater interest in what they learned in their traditional mathe-matics course.

4.4 How Does the Vertical Position of the Point of Inter-section Vary?

We now discuss observation (2).



Looking back at Fig.4.4 it appears that the vertical range of varia-tion of the intersections is rather short. And furthermore, it appearsthat the lowest point of intersection is obtained when the starting pointis the midpoint of the side. The situation where the starting point isthe midpoint is shown in Fig.4.7, and the position of the intersectionpoint can be calculated by the Pythagorean relation and the propor-tionality of corresponding sides of similar triangles. Letting the sides

of the square be of length 1, the point of intersection is found to be3

8from the lower edge.

The highest point of intersection is obtained when the starting pointis an endpoint of the edge (Fig.4.10). Here the point of intersection

coincides with the center of the square, that is,1

2from the lower edge.

Fig. 4.7 The position of the intersection of X-creases by the First Theorem fold.

These values may be obtained more precisely by representing theX-creases as equations and solving the system. Let the x and y-axes betwo sides of the square and let a be the distance of the starting pointfrom the y-axis, as in Fig.4.8.

In order to obtain the equation of the creases we need to express thecoordinates of the endpoints in terms of a. The discussion for Fig.1.7 ofTopic 1 may be helpful.

Equation (i) refers to the crease obtained by folding the lower leftvertex onto the starting point; equation (ii) refers to the crease obtained



by folding the lower right vertex onto the starting point.

y = ax + a2

2+

1

2(i)

y = (1 a)x + a2

2. (ii)

Solving simultaneously to obtain the point of intersection, we obtain

ax + a2

2+

1

2= (1 a)x + a

2

2.

Therefore x =1

2. Substitute this value of x in equation (ii) to obtain

y =1

2

(a 1

2

)2+

3

8.

Since 0 a 1, then 38 y 1

2.

Thus the range of variation for the point of intersection is1

2 3

8=

1

8, which proves the result of Figs. 4.7 and 4.8.

Incidentally, the value of x obtained here confirms our earlier state-ment that the points of intersection lie on the midline.

Fig. 4.8 Calculating the range of variation for the point of intersection by thinking X-creases as a graph.

Comment. The above solution was presented by Mrs. Yoko Takamoto,a teacher of the Senior High School attached to Toshimagaoka WomensEducational Institution, while a trainee at Tsukuba Universitys Ex-tension course entitled Origami and Education.



Fig. 4.9 The length FH (= IG) is an interesting fact.

4.5 Wonders Still Continue

Let us now study the lengths of the segments FH and IG, the segmentson the sides of the square intercepted by the X-creases (Fig.4.9). Theseare sides of triangles FJH and IJG respectively. Since J lies on themidline then FJ = JG and HJ = JI. Also, since they are vertical angles,FJH = GJI. Therefore 4FJH 4GJI, so FH = IG.

Using the pieces of square paper that you folded for different initialpoints match the lengths of these segments. It comes out that they areidentical in length, even as the position of E is allowed to change. Howfascinating!

We state this observation:

(4) The segments on the sides of the square intercepted by theX-creases have a fixed length, regardless of the chosen position of theinitial point.

Be reminded that this statement is the result of comparing withpieces of paper, so at this time it would be more accurate to add thephrase seems to be, or highly possible. But for a while let us leaveit as is.

If all segments FG and IG, are equal, then what is their length?Since the selection of the initial point does not affect the length, wecan select a particular one. Select an endpoint as the initial point as in



Fig. 4.10 When the endpoint is chosen as the arbitrary point, it is obvious IG =1

2.

Fig.4.10. Judging from the figure one can easily understand that thesegment FH is one half of the side of the square.

Therefore on the assumption of (4), we have arrived at:

(5) The lengths of the segments intercepted on the sides of the

square by the X-creases are1

2of the side of the square.

4.6 Solving the Riddle of 1

2

Observation (4) still needs to be proven. We use Fig.4.9. The state-ment can be proved if it can be shown that triangles BEC and FJH aresimilar, and that the height of 4FJH is half that of 4BEC.

With this in mind, first look at 4BAE and 4BMF. Since both tri-angles are right and have a common angle, they are similar. ThereforeBEA = BFM (or HFJ).

Likewise, on the right side, look at 4CED and 4CIN. They are alsosimilar, so CED = CIN (or GIJ).

Finally look at triangles BEC and FJH. Since BEA and EBC arealternate interior angles of parallel edges, they are equal. ThereforeEBC = JFH. Likewise, ECB = JHF. Therefore triangles BECand FJH are similar. But the height of the larger triangle is 1 and the



height of the smaller is1

2(since J is on the midline), therefore the ratio

of their heights, as also corresponding sides is 1 : 2. In particular, FH

is1

2of a side.

4.7 Another Wonder

If X-creases are constructed not only for squares but also for rectan-gles, would statements (1) to (5) hold? Let us study this matter by ex-perimenting with different shapes of rectangles. Include a half-square(that is, cut a square along a midline), and a size A4 paper. For thehalf-square the ratio of the sides is 1 : 2, while for the A4 paper theratio of the sides is 1 :

2.

The position of the rectangle should be such that the shorter sidesare the upper and lower edges; otherwise in the folding the creases willbe forced off the paper through the lower edge instead of through theleft and right edges.

The result of your experimenting should lead you to believe thatstatements (1), (2), (3) and (4) apply to all rectangles. The mathemati-cal proof is left to the reader.

From your experiments, notice that as the rectangles become nar-rower and longer the X-creases become flatter, and the intercepts onthe sides become shorter. Therefore, for rectangles statement (5) willneed to be revised. Fig.4.11 helps us visualize the problem for the gen-eral rectangle with dimensions a and b. Calculating the length of FH interms of a and b is straightforward and is left of the reader. The tablefollowing summarizes the results for different sizes of rectangles.



Fig. 4.11 Generalization of the mathematical principles of X-creases.

1 1

8

1

2

1

16

1

2

1

32

Size of Paper Ratio of short side to long side

Range of variation of

y-coordinate of intersection

point

Square

A4 paper

of long side

of long side

of long side Half-square

The outcomes in the last column are fascinating; do you see thegeometric sequence?


Topic 5

INTRASQUARES ANDEXTRASQUARES

Drawing the Map of the Distribution of the Position of the ReferenceVertex According to the Shape of the Flap Formed by Folding a Piece of

Paper Once

Most of the subject matter that has been taken up so far had as prereq-uisite knowledges the Pythagorean Theorem, quadratic equations, in-center, circumcenter, and excenter of triangles, and therefore are moresuited to third year middle school (9th grade) students and above. Butthere have been requests for topics within the grasp of first and sec-ond year middle school (7th and 8th grade) students, so here is a topicwhich could foster logical thinking that does not rely on too many the-orems and equations.

Comment on Teaching. I have tried using this lesson in lectures tothird year middle school (9th grade) students and in popular lectures,and even to fourth to sixth grade elementary school pupils and firstand second year middle school (7th and 8th grade) students. While itis an effort to cram the whole lecture into a 45 to 50-minute period, itis possible to divide the lecture into two parts: Inner Subdivision andOuter Subdivision (these terms shall be explained later); or conductthe lesson during more flexible time slots, or do it as a group activity.

Here now is the lesson, presented as a sequence of instructions to agroup of learners.

45


Drawing the Map of the Distribution of the Position 46

5.1 Do Not Fold Exactly into Halves

An origami square of any size or color will do. A square cut from ordi-nary bond paper will do, but it would be easier for the teacher to giveinstructions and the pupils to make observations if the two sides canbe distinguished - say one side is colored. Distribute four or five sheetsfor each pupil.

In making an ordinary origami object (e.g., a crane), the first fold iseither the book fold or the diagonal fold, each one dividing the squareinto two equal parts. But for the present topic, the square is not dividedinto two equal parts. Make sure the pupils are aware of this fact beforegiving the following instructions.

Put one sheet on the table. With the colored side facing down andthe white side up, fold only once. Do not fold the paper into two equalparts. In other words, you are not to put a corner or an edge exactly onanother. You are to make a random fold. You may fold the paper at anyplace or in any direction, but fold it neatly and carefully.

At first the pupils might be hesitant and may look around for en-couragement. Remind them, Fold only once.

After the pupils are done folding once, ask them to raise their workso everyone else can see them. There will be a variety of folds. Somemay make a short fold that cuts off only a small portion of the paper(Fig.5.1(e)), while others may make a long fold to cut off a larger portion(Fig.5.1(d) and (k)).

5.2 What Kind of Polygons Can You Get?

Some of you made short folds while others made long ones, but thelength of the fold does not matter. What we are interested in is theshape of the colored flap that you see after you folded the paper. If ithas three corners then it is a triangle; if it has four then it is a quadri-lateral.

The pupils should find it easy to identify the shapes. Ask those withtriangles to raise their pieces of paper; then also those with quadrilat-


47 INTRASQUARES AND EXTRASQUARES

Fig. 5.1 What kinds of polygons can you get?

erals. Ask if other shapes were obtained. Some pupils may come upwith five-sided figures (pentagons); if there are, tell them, Good. Wewill study that later, so please save your work. If some report six ormore corners, they could be looking at the whole paper and not just thecolored portion, so call their attention to this.

Figure 5.1 shows examples of folds that result in triangular flaps((a) to (f)) and quadrilateral flaps ((g) to (l)).

Comment. In my lecture at a third year middle school (9th grade) forgirls in Tokyo, of the 123 pupils the number who came up with triangleswas almost the same as the number who came up with quadrilaterals.In a special lecture given to 150 pupils in the higher grades at an el-ementary school in Yamagata Prefecture, two fifths had triangles andabout three fifths had quadrilaterals. In a lecture at a middle school inIbaraki Prefecture participated in by 80 first year middle school pupils



and 40 guardians, three fifths formed triangles. In an origamics work-shop mainly for teachers but with about 40 nonteachers attending, twothirds had triangles while only one third had quadrilaterals. One cameup with a five-sided polygon. The size of the paper and the mannerby which the folding was demonstrated may account for the differencesmentioned.

Let us proceed to the next step.

Ask the pupils to bring out another piece of paper. Then instructthem: Now those who had triangles previously, make a fold to form aquadrilateral flap; those who obtained quadrilaterals previously makea fold to form a triangular flap. This way each pupil would have thetwo shapes.

5.3 How do You Get a Triangle or a Quadrilateral?

Ask the pupil to examine the two kinds of flaps that they made andcompare the manner by which they were folded. If, by just using theirown pieces of folded paper as examples they find it difficult to figureout conditions by which a triangle or quadrilateral is formed, makedifferent folded flaps as shown in Fig.5.1. Pupils tend to think thatthere is only one correct answer, so remind them there are manydifferent valid ideas.

After some time, expressions on some faces would indicate that theyhave somehow figured out something. Ask them to tell what they foundout.

We can expect to hear different ideas from them, here are somepossibilities.

Mr. A: Open the folded paper and examine the line of the fold. If theline connects two adjacent sides of the original origami square,then the flap is a triangle; if the line connects two oppositesides, then the flap is a quadrilateral.

Ms. B: If in folding only one vertex of the original square moves, then


49 INTRASQUARES AND EXTRASQUARES

the resulting flap is a triangle; if two vertices move, then theflap is a quadrilateral.

Mr. C: The shape of the flap depends on the position of the moved ver-tex on the origami paper.

Ms. D: If the colored portion is contained wholly in the origami square,then it is a triangle; if a part of it is outside the square then theflap is a quadrilateral.

Mr. E: If the line of the fold passes through any of the vertices of thesquare, then surely the resulting flap is a triangle. If the lineof fold does not pass through any vertex then the flap could beeither a triangle or a quadrilateral.

Ms. F: If the line of fold is shorter than a side of the square, then theflap is a triangle; if longer, then we could have either of the two.

These are just some of the many possible discoveries that the pupilsmay make. Tackle each one of them by asking the others what theythink of the comment. For example, one may say that Mr. As idea isnot applicable when the line of fold hits a vertex. As for Ms. Ds idea,a counterexample can be found by folding parallel to a side (Fig.5.1(g)).- that is, a quadrilateral may be formed even if the colored flap fallsentirely on the paper square. It is important that when leading thediscussion, one should not judge right away that an idea is right orwrong For example, one should not say that Mr. Xs answer is corrector Ms. Ys is wrong. Each person has discovered something, althoughan idea may work only in certain situations. The teacher can clarifyby identifying the conditions or cases where the idea works. The ideasgiven by the pupils should help the teacher in deciding what to examineor explore next.

5.4 Now to Making a Map

We continue looking for conditions that lead to the formation of a trian-gular or a quadrilateral flap. One way is to pursue the line of thinking



of one of the pupils. Ms. B and Mr. C focused on the movement of onevertex; let us pursue Mr. Cs idea. Use a new sheet of square paper.Mark at random about ten points on the paper. Select one vertex of thepaper square to be the reference vertex (in Fig.5.2 the reference vertexis the lower left vertex, marked ). Then carefully move it to a markedpoint and determine the shape of the flap formed. Write 3 or 4 be-side the point or color it, to indicate the number of the vertices of thepolygonal flap.

Fig. 5.2 Investigating what kind of polygons we can get by plotting points randomly.

Do this for the other marked points. As more and more of the ran-dom points are labeled, a map begins to be formed. Notice how the3s cluster in certain areas of the paper square; so also the 4s. Ifone thinks of the square as a village divided into lots, then the clustersof 3s may be called Lot 3 while the clusters of 4s may be calledLot 4.

In order to better identify the boundaries of Lot 3 and Lot 4, letus repeat the experiment, but this time select points more systemat-ically (if your paper is messy, use a new sheet). Using a ruler drawequally-spaced vertical and horizontal lines to make a grid. For a pie

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